Question

In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80^o and angle ACE = 10^o. Calculate : Angle BCD

Answer 1

DC || EB

∴ DCE = ∠BEC = 50° (Alternate angles)

∴ ∠ AOB = 80°

⇒ ∠ ACB = `1 /2` ∠AOB = 40°

(Angle at the center is double the angle at the circumference subtended by the same chord) We have,

∠BCD = ∠ACB + ∠ACE + ∠DCE = 40° +10°+ 50° = 100°